Let $G$ be finite and every nonidentity element have prime order. If $Z(G)\neq\{e\}$, prove that every nonidentity element of $G$ has the same order.

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I'm reading "Contemporary Abstract Algebra," by Gallian.

This is Exercise 4.51.$$^\dagger$$

Suppose that $$G$$ is a finite group with the property that every nonidentity element has prime order (e.g., $$D_3$$). If $$Z(G)$$ is not trivial, prove that every nonidentity element of $$G$$ has the same order.

Thoughts:

Lemma: If $$G$$ is abelian with the property that every nonidentity element has prime order, then every nonidentity element has the same prime order.

Proof: If $$G$$ is abelian (i.e., $$G=Z(G)$$), then consider $$g,h\in G$$ such that $$\lvert g\rvert=p$$ and $$\lvert h\rvert=q$$ for distinct primes $$p$$ and $$q$$. We have

\begin{align} (gh)^{pq}&=(g^p)^q(h^q)^p\\ &=e, \end{align}

so that $$\lvert gh\rvert$$ divides $$pq$$.

If $$\lvert gh\rvert=pq$$, then it is composite, a contradiction; thus without loss of generality $$\lvert gh\rvert=p$$. Now we have

\begin{align} e&=(gh)^p\\ &=g^ph^p\\ &=eh^p \\ &=h^p, \end{align}

but now $$q\mid p$$, which is a contradiction since $$p\neq q$$ and $$p$$ is prime.

Thus all nonidentity elements of $$G$$ have the same prime order.$$\square$$

That's all I have so far.

I've considered proving some version of the contrapositive but nothing springs to mind other than, "yeah . . . contrapositive might work" followed by a shrug.

Edit:

This comment gives me some idea of how to finish; however, I'm not sure where the finiteness of $$G$$ comes into play.

Please help :)

$$\dagger$$ I've just noticed that this exercise has a solution in the book. It makes sense to me. If anyone would like to answer it here anyway, go ahead! I might post an answer later summarising the proof in the text.

• How about mimicking the abelian proof with one general element of the group and one nontrivial element of the center? Commented Jan 16, 2019 at 11:55
• I see what you mean, @Mindlack! That's a fun little trick! Thank you. Commented Jan 16, 2019 at 11:57
• But . . . Why is finiteness necessary, then, in the original exercise? @Mindlack. Commented Jan 16, 2019 at 11:58
• @Shaun They might just be trying to avoid awkward questions like "is infinity prime". Commented Jan 16, 2019 at 12:17
• @Shaun They don't. The problem is that it's kind of an awkward edge-case definition thing, and I wouldn't be surprised if someone just ruled out the infinite case to avoid having to think about it. Commented Jan 16, 2019 at 12:24

2 Answers

The following paraphrases the proof in the solutions section of the book the exercise is from.

Let $$z\in Z(G)$$ such that $$\lvert z\rvert=p$$ for a prime $$p$$. Consider $$g\in G$$. We have $$\lvert g\rvert=q$$ is prime. Then

\begin{align} (zg)^{pq}&=(z^p)^q(g^q)^p \\ &=e^qe^p \\ &=e \end{align}

and thus $$\lvert zg\rvert\in\{p, q\}$$ (since it has to be prime; and if $$\lvert zg\rvert=1$$, then $$z=g^{-1}$$, so $$p=q$$). Assume without loss of generality that $$\lvert zg\rvert=p$$. Then

\begin{align} e&=(zg)^p \\ &=z^pg^p \\ &=eg^p \\ &=g^p, \end{align}

so $$q\mid p$$. Hence $$p=q$$.

• Why can’t $|zg| = 1$, i.e how do you know that $zg \neq e$? Commented Apr 9, 2020 at 18:07
• Since $|zg|$ has to be prime (and $1$ is not prime), @Junglemath. Commented Apr 9, 2020 at 19:33
• Every nonidentity element has prime order. Why can't $zg = e$? Commented Apr 9, 2020 at 19:40
• I see now, @Junglemath. Thank you. I have edited the answer accordingly. Commented Apr 9, 2020 at 19:47

I am doing selfstudy and studying Group Theory, and using Gallian's book. Today I encounter this problem. I spend almost 2 hours on the problems and didn't find a clear path. So I thought to take some help (hint) from MSE, and so I reached at your question. After reading your question I found where my knowledge lack. Actually the part where you showed p and q are distinct by contradiction, here I was not getting contradiction.
Now I have wrote my proof without looking at the back of the book (solution). Please have a look.

Proof: Suppose $$Z(G)$$ is not trivial. Suppose $$x \in Z(G)$$ and $$y \in G$$ are both non-identity elements. Let $$|x|=p$$ and $$|y|=q$$. Assume for the sake of contradiction that $$p$$ and $$q$$ are distinct. Since $$(xy)^{pq} =e$$, it follows $$|xy|$$ divides $$pq$$. If $$|xy|=pq$$ then a contradiction because $$pq$$ is composite. WLOG assume $$|xy|=p$$. Then $$e=(xy)^p=y^p$$, so $$q|p$$, a contradiction. Thus $$p$$ and $$q$$ are equal. Since $$p$$ and $$q$$ were arbitrary, therefore every non-identity element of $$G$$ has same order. ∎

Edit: I just have looked at your accepted answer and solution in the book. I found I have missed a case in which order of $$pq$$ is equal to one.